let A be non empty closed_interval Subset of REAL; for f being Function of A,REAL st f | A is bounded holds
for D, D1 being Division of A ex D2 being Division of A st
( D <= D2 & D1 <= D2 & rng D2 = (rng D1) \/ (rng D) & 0 <= (lower_sum (f,D2)) - (lower_sum (f,D)) & 0 <= (lower_sum (f,D2)) - (lower_sum (f,D1)) )
let f be Function of A,REAL; ( f | A is bounded implies for D, D1 being Division of A ex D2 being Division of A st
( D <= D2 & D1 <= D2 & rng D2 = (rng D1) \/ (rng D) & 0 <= (lower_sum (f,D2)) - (lower_sum (f,D)) & 0 <= (lower_sum (f,D2)) - (lower_sum (f,D1)) ) )
assume A1:
f | A is bounded
; for D, D1 being Division of A ex D2 being Division of A st
( D <= D2 & D1 <= D2 & rng D2 = (rng D1) \/ (rng D) & 0 <= (lower_sum (f,D2)) - (lower_sum (f,D)) & 0 <= (lower_sum (f,D2)) - (lower_sum (f,D1)) )
for D, D1 being Division of A ex D2 being Division of A st
( D <= D2 & D1 <= D2 & rng D2 = (rng D1) \/ (rng D) & 0 <= (lower_sum (f,D2)) - (lower_sum (f,D)) & 0 <= (lower_sum (f,D2)) - (lower_sum (f,D1)) )
proof
let D,
D1 be
Division of
A;
ex D2 being Division of A st
( D <= D2 & D1 <= D2 & rng D2 = (rng D1) \/ (rng D) & 0 <= (lower_sum (f,D2)) - (lower_sum (f,D)) & 0 <= (lower_sum (f,D2)) - (lower_sum (f,D1)) )
consider D2 being
Division of
A such that A6:
D <= D2
and A7:
D1 <= D2
and A8:
rng D2 = (rng D1) \/ (rng D)
by Th4;
A9:
(lower_sum (f,D2)) - (lower_sum (f,D1)) >= 0
by A1, A7, INTEGRA1:46, XREAL_1:48;
(lower_sum (f,D2)) - (lower_sum (f,D)) >= 0
by A1, A6, INTEGRA1:46, XREAL_1:48;
hence
ex
D2 being
Division of
A st
(
D <= D2 &
D1 <= D2 &
rng D2 = (rng D1) \/ (rng D) &
0 <= (lower_sum (f,D2)) - (lower_sum (f,D)) &
0 <= (lower_sum (f,D2)) - (lower_sum (f,D1)) )
by A6, A7, A8, A9;
verum
end;
hence
for D, D1 being Division of A ex D2 being Division of A st
( D <= D2 & D1 <= D2 & rng D2 = (rng D1) \/ (rng D) & 0 <= (lower_sum (f,D2)) - (lower_sum (f,D)) & 0 <= (lower_sum (f,D2)) - (lower_sum (f,D1)) )
; verum